Parabola Graphs A97F82

1. **Problem Statement:** Graph the inequality $y \geq x^2 - 2$ and find the x-intercept, vertex, and test point. 2. **Formula and Rules:** The graph is a parabola of the form $y = ax^2 + bx + c$ with $a=1$, $b=0$, $c=-2$. The vertex formula is $\left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right)$. 3. **Find the vertex:** $$x_v = -\frac{0}{2 \times 1} = 0$$ $$y_v = -2$$ Vertex is at $(0, -2)$. 4. **Find the x-intercepts:** Set $y=0$: $$0 = x^2 - 2$$ $$x^2 = 2$$ $$x = \pm \sqrt{2}$$ 5. **Test point:** Choose $x=0$, then $y = 0^2 - 2 = -2$. Since $y \geq x^2 - 2$, test point $(0,0)$: $$0 \geq 0^2 - 2 \Rightarrow 0 \geq -2$$ True, so shade above the parabola. --- 1. **Problem Statement:** Graph $y < x^2 + 2x - 3$ and find x-intercept, vertex, test point. 2. **Vertex:** $$a=1, b=2, c=-3$$ $$x_v = -\frac{2}{2 \times 1} = -1$$ $$y_v = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4$$ Vertex at $(-1, -4)$. 3. **X-intercepts:** Set $y=0$: $$0 = x^2 + 2x - 3$$ $$x^2 + 2x - 3 = 0$$ $$(x+3)(x-1) = 0$$ $$x = -3, 1$$ 4. **Test point:** Use $(0,0)$: $$0 < 0^2 + 2(0) - 3 = -3$$ $$0 < -3$$ False, so shade below the parabola. --- 1. **Problem Statement:** Graph $x^2 + y \leq 16$. 2. **Rewrite:** $$y \leq 16 - x^2$$ 3. **Vertex:** $$a = -1, b=0, c=16$$ $$x_v = 0$$ $$y_v = 16$$ Vertex at $(0,16)$. 4. **X-intercepts:** Set $y=0$: $$0 = 16 - x^2$$ $$x^2 = 16$$ $$x = \pm 4$$ 5. **Test point:** $(0,0)$: $$0 \leq 16 - 0 = 16$$ True, shade below the parabola. --- 1. **Problem Statement:** Graph $y > -x^2 + 6$. 2. **Vertex:** $$a = -1, b=0, c=6$$ $$x_v = 0$$ $$y_v = 6$$ Vertex at $(0,6)$. 3. **X-intercepts:** Set $y=0$: $$0 = -x^2 + 6$$ $$x^2 = 6$$ $$x = \pm \sqrt{6}$$ 4. **Test point:** $(0,0)$: $$0 > -0 + 6 = 6$$ False, so shade above the parabola. --- 1. **Problem Statement:** Graph $y \leq x^2 - 4x + 1$. 2. **Vertex:** $$a=1, b=-4, c=1$$ $$x_v = -\frac{-4}{2 \times 1} = 2$$ $$y_v = 2^2 - 4(2) + 1 = 4 - 8 + 1 = -3$$ Vertex at $(2, -3)$. 3. **X-intercepts:** Set $y=0$: $$0 = x^2 - 4x + 1$$ Use quadratic formula: $$x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(1)}}{2} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$$ 4. **Test point:** $(0,0)$: $$0 \leq 0^2 - 4(0) + 1 = 1$$ True, shade below the parabola. --- **Final answers:** - Each parabola's vertex and x-intercepts are found. - Test points confirm shading region. - Graphs are parabolas opening up or down with shading as per inequality.